STD 12 - 3 and 4 . determinant and metrices — Mathematics STD 12 Science — Question

$A=\left[\begin{array}{ll}5 & 2 \\ 7 & 3\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and  $B=\left[\begin{array}{l}4 \\ 5\end{array}\right]$

Now $|A|=15-14=1 \neq 0$

Thus, $A$ is non-singular. Therefore, its inverse exists.

Now,

$A^{-1}=\frac{1}{|\mathrm{A}|}(\operatorname{adj} A)$

$\therefore A^{-1}=\left[\begin{array}{cc}3 & -2 \\ -7 & 5\end{array}\right]$

$\therefore X=A^{-1} B=\left[\begin{array}{cc}3 & -2 \\ -7 & 5\end{array}\right]\left[\begin{array}{l}4 \\ 5\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}12-10 \\ -28+25\end{array}\right]=\left[\begin{array}{c}2 \\ -3\end{array}\right]$

Hence, $x=2$ and $y=-3$